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- 1. 18Three friends divided some Find sum of digits of D.bullets equally. After all of themshot 4 bullets the total number Letof bullets remaining is equal to A= 19991999the bullets each had after B = sum of digits of Adivision. Find the original C = sum of digits of Bnumber divided. D = sum of digits of CAnswer (HINT : A = B = C = D (mod 9))18 AnswerAssume that initial there were The sum of the digits od D is3*X bullets. 1.So they got X bullets each after Let E = sum of digits of D.division. It follows from the hint that A =All of them shot 4 bullets. So E (mod 9)now they have (X - 4) bulletseach.But it is given that,after theyshot 4 bullets each, totalnumber of bullets remaining isequal to the bullets each hadafter division i.e. XTherefore, the equation is3 * (X - 4) = X3 * X - 12 = X2 * X = 12X=6Therefore the total bulletsbefore division is = 3 * X = 18 1
- 2. Consider, so 19991999 = 1 A = 19991999 (mod 9) < 20002000 Therefore we conclude that = 22000 * E=1.10002000 There is a 50m long army platoon marching ahead. The last person in = 1024200 * the platoon wants to give a letter to106000 the first person leading the platoon. So while the platoon is < 10800 * 106000 marching he runs ahead, reaches the first person and hands over the = 106800 letter to him and without stopping he runs and comes back to his original position. i.e. A < 106800 In the mean time the whole platoon has moved ahead by 50m. i.e. B < 6800 *9 = 61200 The question is how much distance did the last person cover in that i.e. C < 5 * 9 time. Assuming that he ran the= 45 whole distance with uniform speed. i.e. D < 2 * 9 Submitted= 18 Answer i.e. E <= 9 The last person covered 120.71 meters. i.e. E is asingle digit number. It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount Also, of time. Hence, the ratio of the 1999 = 1 (mod distance they covered - while9) person moving forward and backword - are equal. 2
- 3. Lets assume that when the lastperson reached the first person, If you take a marker & startthe platoon moved X meters from a corner on a cube, whatforward. is the maximum number of edges you can trace across ifThus, while moving forward the you never trace across thelast person moved (50+X) same edge twice, nevermeters whereas the platoon remove the marker from themoved X meters. cube, & never trace anywhere on the cube, except for theSimilarly, while moving back corners & edges?the last person moved [50-(50-X)] X meters whereas the Answerplatoon moved (50-X) meters. 9Now, as the ratios are equal,(50+X)/X = X/(50-X) To verify this, you can make a(50+X)*(50-X) = X*X drawing of a cube, & number each of its 12 edges. Then,Solving, X=35.355 meters always starting from 1 corner & 1 edge, you can determine allThus, total distance covered by of the possible combinations forthe last person tracing along the edges of a= (50+X) + X cube.= 2*X + 50= 2*(35.355) + 50 There is no need to start from= 120.71 meters other corners or edges of the cube, as you will only be repeating the sameNote that at first glance, one combinations. The process is amight think that the total little more involved than this,distance covered by the last but is useful for solving manyperson is 100 meters, as he ran types of spatial puzzles.the total lenght of the platoon(50 meters) twice. TRUE, but One of Mr. Bajaj, his wife, theirthats the relative distance son and Mr. Bajajs mother iscovered by the last person i.e. an Engineer and another is aassuming that the platoon is Doctor.stationary. 3
- 4. If the Doctor is a male, Doctor is either his wife or his then the Engineer is a son . It is not possible to male. determine anything further. If the Engineer is younger than the Doctor, then the Engineer and the Doctor Three men - Sam, Cam and are not blood relatives. Laurie - are married to Carrie, If the Engineer is a Billy and Tina, but not female, then she and the necessarily in the same order. Doctor are blood relatives. Sams wife and Billys Husband play Carrie and Tinas husbandCan you tell who is the Doctor at bridge. No wife partners herand the Engineer? husband and Cam does notAnswer play bridge.Mr. Bajaj is the Engineer and Who is married to Cam?either his wife or his son isthe Doctor. AnswerMr. Bajajs wife and mother arenot blood relatives. So from 3, if Carrie is married to Cam.the Engineer is a female, theDoctor is a male. But from 1, if "Sams wife and Billysthe Doctor is a male, then the Husband play Carrie and TinasEngineer is a male. Thus, there husband at bridge."is a contradiction, if theEngineer is a female. Hence, It means that Sam is noteither Mr. Bajaj or his son is the married to either Billy or Carrie.Engineer. Thus, Sam is married to Tina.Mr. Bajajs son is the youngest As Cam does not play bridge,of all four and is blood relative Billys husband must be Laurie.of each of them. So from 2, Mr.Bajajs son is not the Engineer. Hence, Carrie is married toHence, Mr. Bajaj is the Cam.Engineer.Now from 2, Mr. Bajajs mother There are 3 persons X, Y andcan not be the Doctor. So the Z. On some day, X lent tractors 4
- 5. to Y and Z as many as they tractors, Y had 21 tractorshad. After a month Y gave as and Z had 12 tractors.many tractors to X and Z asmany as they have. After a A certain street has 1000month Z did the same thing. At buildings. A sign-maker isthe end of this transaction each contracted to number theone of them had 24. houses from 1 to 1000. How many zeroes will he need?Find the tractors each originallyhad? AnswerAnswer The sign-maker will need 192 zeroes.One way to solve it is bymaking 3 equations and solve Divide 1000 building numbersthem simultaneously. But there into groups of 100 each asis rather easier way to solve it follow:using Backtracing. (1..100), (101..200), (201..300), ....... (901..1000)Its given that at the end, eachhad 24 tractors (24, 24, 24) i.e. For the first group, sign-makerafter Z gave tractors to X & Y will need 11 zeroes.as many as they had. It means For group numbers 2 to 9, hethat after getting tractors from Z will require 20 zeroes each.their tractors got doubled. So And for group number 10, hebefore Z gave them tractors, will require 21 zeroes.they had 12 tractors each and Zhad 48 tractors. (12, 12, 48) The total numbers of zeroes required areSimilarly, before Y gave = 11 + 8*20 + 21tractors to X & Z, they had 6 & = 11 + 160 + 2124 tractors respectively and Y = 192had 42 tractors i.e. (6, 42, 24)Again, before X gave tractors to There are 9 coins. Out of whichY & Z, they had 21 & 12 one is odd one i.e weight is lesstractors respectively and X had or more. How many iterations39 tractors i.e. (39, 21, 12) of weighing are required to find odd coin?Hence, initially X had 39 5
- 6. Answer If L2 is light, L2 is the oddIt is always possible to find odd coin and iscoin in 3 weighings and to tell lighter.whether the odd coin is heavier If L3 is light,or lighter. L3 is the odd coin and is 1. Take 8 coins and weigh 4 lighter. against 4. o If both are not equal, goto step 2 o If both are equal, o If (H1, H2, L1) is goto step 3 heavier side on the scale, either H1 or H2 is heavier. Weight H1 against 2. One of these 8 coins is H2 the odd one. Name the If both are coins on heavier side of equal, there is the scale as H1, H2, H3 some error. and H4. Similarly, name If H1 is heavy, the coins on the lighter H1 is the odd side of the scale as L1, coin and is L2, L3 and L4. Either one heavier. of Hs is heavier or one of If H2 is heavy, Ls is lighter. Weigh (H1, H2 is the odd H2, L1) against (H3, H4, coin and is X) where X is one coin heavier. remaining in intial weighing. o If both are equal, o If (H3, H4, X) is one of L2, L3, L4 is heavier side on the lighter. Weigh L2 scale, either H3 or against L3. H4 is heavier or L1 If both are is lighter. Weight H3 equal, L4 is against H4 the odd coin If both are and is lighter. equal, L1 is 6
- 7. the odd coin 3. On the nth and last day, and is lighter. the remaining n medals If H3 is heavy, were awarded. H3 is the odd coin and is How many days did the contest heavier. last, and how many medals If H4 is heavy, were awarded altogether? H4 is the odd Answer coin and is heavier. Total 36 medals were awarded and the contest was for 6 days. 3. The remaining coin X is On day 1: Medals awarded = (1 the odd one. Weigh X + 35/7) = 6 : Remaining 30 against the anyone coin medals used in initial weighing. On day 2: Medals awarded = (2 o If both are equal, + 28/7) = 6 : Remaining 24 there is some error. medals o If X is heavy, X is On day 3: Medals awarded = (3 the odd coin and is + 21/7) = 6 : Remaining 18 heavier. medals o If X is light, X is the On day 4: Medals awarded = (4 odd coin and is + 14/7) = 6 : Remaining 12 lighter. medals On day 5: Medals awarded = (5In a sports contest there were +7/7) = 6 : Remaining 6 medalsm medals awarded on n On day 6: Medals awarded 6successive days (n > 1). I got this answer by writing 1. On the first day 1 medal small program. If anyone know and 1/7 of the remaining any other simpler method, do m - 1 medals were submit it. awarded. 2. On the second day 2 A number of 9 digits has the medals and 1/7 of the following properties: now remaining medals was awarded; and so on. The number comprising the leftmost two digits is divisible by 2, that 7
- 8. comprising the leftmost container evaporated on the 1st three digits is divisible by day. 3/4th of the remaining 3, the leftmost four by 4, contents of the container the leftmost five by 5, and evaporated on the second day. so on for the nine digits of the number i.e. the What part of the contents of the number formed from the container is left at the end of first n digits is divisible by the second day? n, 2<=n<=9. Each digit in the number is different i.e. no digits are repeated. Answer The digit 0 does not occur in the number i.e. it is Assume that contents of the comprised only of the container is X digits 1-9 in some order. On the first day 1/3rd isFind the number. evaporated. (1 - 1/3) of X is remaining i.e.Answer (2/3)XThe answer is 381654729 On the Second day 3/4th is evaporated. Hence,One way to solve it is Trial-&- (1- 3/4) of (2/3)X is remainingError. You can make it bit i.e. (1/4)(2/3)X = (1/6) Xeasier as odd positions willalways occupy ODD numbers Hence 1/6th of the contents ofand even positions will always the container is remainingoccupy EVEN numbers. Further5th position will contain 5 as 0does not occur. Vipul was studying for his examinations and the lightsThe other way to solve this went off. It was around 1:00problem is by writing a AM. He lighted two uniformcomputer program that candles of equal length but onesystematically tries all thicker than the other. The thickpossibilities candle is supposed to last six hours and the thin one two1/3 rd of the contents of a hours less. When he finally went to sleep, the thick candle 8
- 9. was twice as long as the thin If you started a business inone. which you earned Rs.1 on the first day, Rs.3 on the secondFor how long did Vipul study in day, Rs.5 on the third day, Rs.7candle light? on the fourth day, & so on.Answer How much would you have earned with this business afterVipul studied for 3 hours in 50 years (assuming there arecandle light. exactly 365 days in every year)?Assume that the initial lenght ofboth the candle was L and AnswerVipul studied for X hours. Rs.333,062,500In X hours, total thick candleburnt = XL/6 To begin with, you want toIn X hours, total thin candle know the total number of days:burnt = XL/4 365 x 50 = 18250.After X hours, total thick candle By experimentation, theremaining = L - XL/6 following formula can beAfter X hours, total thin candle discovered, & used toremaining = L - XL/4 determine the amount earned for any particular day: 1 + 2(x-Also, it is given that the thick 1), with x being the number ofcandle was twice as long as the the day. Take half of the 18250thin one when he finally went to days, & pair them up with thesleep. other half in the following way:(L - XL/6) = 2(L - XL/4) day 1 with day 18250, day 2(6 - X)/6 = (4 - X)/2 with day 18249, & so on, & you(6 - X) = 3*(4 - X) will see that if you add these6 - X = 12 - 3X pairs together, they always2X = 6 equal Rs.36500.X=3 Multiply this number by the totalHence, Vipul studied for 3 number of pairs (9125), & youhours i.e. 180 minutes in candle have the amount you wouldlight. have earned in 50 years. 9
- 10. Math gurus may use series It is given that the time betweenformula to solve it.(series: first and last ticks at 6o is 301,3,5,7,9,11.....upto 18250 seconds.terms) Total time gaps between firstA worker earns a 5% raise. A and last ticks at 6o = 5year later, the worker receives (i.e. between 1 & 2, 2 & 3, 3 &a 2.5% cut in pay, & now his 4, 4 & 5 and 5 & 6)salary is Rs. 22702.68 So time gap between two ticksWhat was his salary to begin = 30/5 = 6 seconds.with?Answer Now, total time gaps betweenRs.22176 first and last ticks at 12o = 11 Therefore time taken for 12Assume his salary was Rs. X ticks = 11 * 6 = 66 seconds (and not 60 seconds)He earns 5% raise. So hissalary is (105*X)/100 500 men are arranged in anA year later he receives 2.5% array of 10 rows and 50cut. So his salary is columns according to their((105*X)/100)*(97.5/100) which heights.is Rs. 22702.68 Tallest among each row of allHence, solving equation are asked to come out. And the((105*X)/100)*(97.5/100) = shortest among them is A.22702.68X = 22176 Similarly after resuming them toAt 6o a clock ticks 6 times. The their original positions, thetime between first and last ticks shortest among each columnis 30 seconds. How long does it are asked to come out. And thetick at 12o. tallest among them is B. Now who is taller A or B ? AnswerAnswer66 seconds 10
- 11. daughter-in-law. How many members are there in Mr. Mehtas family? Give minimal possible answer. Answer There are 7 members in Mr. Mehtas family. Mother & Father of Mr. Mehta, Mr. & Mrs. Mehta,No one is taller, both are his son and two daughters.same as A and B are thesame person. Mother & Father of Mr. Mehta |As it is mentioned that 500 men |are arranged in an array of 10 Mr. & Mrs. Mehtarows and 50 columns |according to their heights. |Lets assume that position One Son & Two Daughtersnumbers represent their When Alexander the Greatheights. Hence, the shortest attacked the forces of Porus, anamong the 50, 100, 150, ... Indian soldier was captured by450, 500 is person with height the Greeks. He had displayed50 i.e. A. Similarly the tallest such bravery in battle, however,among 1, 2, 3, 4, 5, ..... 48, 48, that the enemy offered to let50 is person with height 50 i.e. him choose how he wanted toB be killed. They told him, "If you tell a lie, you will put to theNow, both A and B are the sword, and if you tell the truthperson with height 50. Hence you will be hanged."both are same. The soldier could make onlyIn Mr. Mehtas family, there are one statement. He made thatone grandfather, one statement and went free. Whatgrandmother, two fathers, two did he say?mothers, one father-in-law, onemother-in-law, four children, Answerthree grandchildren, onebrother, two sisters, two sons, The soldier said, "You will puttwo daughters and one me to the sword." 11
- 12. mistake and gave him Y rupeesThe soldier has to say a and X paise. Neither the personParadox to save himself. If his nor the cashier noticed that.statement is true, he will behanged, which is not the sword After spending 20 paise, theand hence false. If his person counts the money. Andstatement is false, he will be to his surprise, he has doubleput to the sword, which will the amount he wanted tomake it true. A Paradox !!! withdraw.A person wanted to withdraw X Find X and Y. (1 Rupee = 100rupees and Y paise from the Paise)bank. But cashier made a 2 * (100X + Y) = 100Y + X - 20 200X + 2Y =Answer 100Y +X - 20As given, the person wanted to 199X - 98Y = -withdraw 100X + Y paise. 20But he got 100Y + X paise. 98Y - 199X = 20After spending 20 paise, he hasdouble the amount he wanted Now, we got one equation; butto withdraw. Hence, the there are 2 variables. We haveequation is to apply little bit of logic over here. We know that if we interchange X & Y, amount gets double. So Y should be twice of X or one more than twice of X i.e. Y = 2X or Y = 2X+1 Case I : Y=2X Solving two equations simultaneously 12
- 13. 98Y - 199X = 20Y - 2X = 0We get X = - 20/3 & Y = - 40/2 Who will win the game, O or X? Can you tell which was theCase II : Y=2X+1 sixth mark and at whichSolving two equations position? Do explain yoursimultaneously answer.98Y - 199X = 20 At the Party:Y - 2X = 1We get X = 26 & Y = 53 1. There were 9 men and children.Now, its obvious that he wanted 2. There were 2 moreto withdraw Rs. 26.53 women than children. 3. The number of differentThe game of Tic-Tac-Toe is man-woman couplesbeing played between two possible was 24. Noteplayers. Only the last mark to that if there were 7 menbe placed in the game as and 5 women, then thereshown. would have been 35 man- woman couples possible. Also, of the three groups - men, women and children - at the party: 4. There were 4 of one group. 5. There were 6 of one group. 6. There were 8 of one group. Exactly one of the above 6 statements is false. Can you tell which one is false? Also, how many men, women and children are there at the party 13
- 14. Assume that both the players Assume that Statements (4),are intelligent enough. (5) and (6) are all true. Then, Statement (1) is false. But then Statement (2) and (3) both can not be true. Thus, contradictory to the fact that exactly oneAnswer statement is false.O will win the game. The So Statement (4) or Statementsixth mark was X in square 9. (5) or Statement (6) is false. Also, Statements (1), (2) andThe 7th mark must be placed in (3) all are true.square 5 which is the winsituation for both X and O. From (1) and (2), there are 11Hence, the 6th mark must be men and women. Then fromplaced in a line already (3), there are 2 possible cases -containing two of the opponents either there are 8 men and 3marks. There are two such women or there are 3 men andpossibilities - the 6th mark 8 women.would have been either O insquare 7 or X in square 9. If there are 8 men and 3 women, then there is 1 child.As we know both the players Then Statements (4) and (5)are intelligent enough, the 6th both are false, which is notmark could not be O in square possible.7. Instead, he would haveplaced O in square 5 and would Hence, there are 3 men, 8have won. women and 6 children. Statement (4) is false.Hence, the sixth mark must beX placed in square 9. And the There is a shortage ofseventh mark will be O. Thus O tubelights, bulbs and fans in awill win the game. village - Kharghar. It is found thatAnswer All houses do not haveStatement (4) is false. There either tubelight or bulb orare 3 men, 8 women and 6 fan.children. 14
- 15. exactly 19% of houses do not have all 3 items - tubelight, not have just one of bulb and fan. these. atleast 67% of houses do Thus, 42% houses do not have not have tubelights. tubelight, bulb and fan. atleast 83% of houses do Mr. Subramaniam rents a not have bulbs. private car for Andheri-Colaba- atleast 73% of houses do Andheri trip. It costs him Rs. not have fans. 300 everyday.What percentage of houses do Mr. Subramaniam that therenot have tubelight, bulb and One day the car driver informedfan? were two students from BandraAnswer who wished to go from Bandra to Colaba and back to Bandra.42% houses do not have Bandra is halfway betweentubelight, bulb and fan. Andheri and Colaba. Mr. Subramaniam asked the driverLets assume that there are 100 to let the students travel withhouses. Hence, there should be him.total 300 items i.e. 100tubelights, 100 bulbs and 100 On the first day when theyfans. came, Mr. Subramaniam said, "If you tell me theFrom the given data, we know mathematically correct pricethat there is shortage of atleast you should pay individually for(67+83+73) 223 items in every your portion of the trip, I will let100 houses. you travel for free."Also, exactly 19 houses do not How much should the individualhave just one item. It means student pay for their journey?that remaining 81 houses Answershould account for the shortageof remaining (223-19) 204 The individual student shoulditems. If those remaining 81 pay Rs. 50 for their journey.houses do not have 2 itemseach, there would be a Note that 3 persons areshortage of 162 items. But total travelling between Bandra andof 204 items are short. Hence, Colaba.atleast (204-162) 42 houses do 15
- 16. The entire trip costs Rs. 300 to O UMr. Subramanian. Hence, half Tof the trip costs Rs. 150. ------------For Andheri-Bandra-Andheri, -only one person i.e. Mr.Subramaniam is travelling. S T E M | D E M I SHence, he would pay Rs. 150. EFor Bandra-Colaba-Bandra, | D M O Cthree persons i.e Mr.Subramaniam and two ------------students, are travelling. Hence, -each student would pay Rs. 50. T U I SSubstitute digits for the lettersto make the following Division S T E Mtrue --------- - Z Z Z E Z U M M ------- - I S T Note that the leftmost letter cant be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter M, no other letter can be 3 and all other M in the puzzle must be 16
- 17. Answer O U TC=0, U=1, S=2, T=3, O=4, 4 1 3M=5, I=6, Z=7, E=8, D=9 ------------It is obvious that U=1 (as - ------U*STEM=STEM) and C=0 (as -------I-C=I). S T E M | D E M I SS*O is a single digit and also E 2 3 8 5 | 9 8S*T is a single digit. Hence, 5 6 2 8their values (O, S, T) must be2, 3 or 4 (as they can not be 0 | D M O Cor 1 or greater than 4). | 9 5 4 0Consider, STEM*O=DMOC, ------------where C=0. It means that M - ------must be 5. Now, its simple. -------O=4, S=2, T=3, E=8, Z=7, I=6and D=9. T U I S 3 1 6 2 S T E M 2 3 8 5 --------- - --- ------- Z Z Z E 7 7 7 8 Z U M M 7 1 5 5 ------- - - ------- 17
- 18. 6 * X = 120 + X/2 I S 12 * X = 240 + XT 11 * X = 2406 2 3 X = 21.8182Also, when arranged from 0 to X = 21 minutes 49.5 seconds9, it spells CUSTOMIZED. Hence, at 4:21:49.5 minute hand is exactly aligned with theAt what time after 4.00 p.m. is hour hand.the minutes hand of a clockexactly aligned with the hour A soldier looses his way in ahand? thick jungle. At random he walks from his camp butAnswer mathematically in an interesting fashion.4:21:49.5 First he walks one mile EastAssume that X minutes after then half mile to North. Then4.00 PM minute hand exactly 1/4 mile to West, then 1/8 milealigns with and hour hand. to South and so on making a loop.For every minute, minute handtravels 6 degrees. Finally how far he is from hisHence, for X minutes it will camp and in which direction?travel 6 * X degrees. AnswerFor every minute, hour handtravels 1/2 degrees. The soldier is 0.8944 milesHence, for X minutes it will away from his camp towardstravel X/2 degrees. East-North.At 4.00 PM, the angle between It is obvious that he is in East-minute hand and hour hand is North direction.120 degrees. Also, after Xminutes, minute hand and hour Distance travelled in North andhand are exactly aligned. So South directionsthe angle with respect to 12 i.e. = 1/2 - 1/8 + 1/32 - 1/128 +Vertical Plane will be same. 1/512 - 1/2048 + and so on... (aTherefore, geometric series with r = (-1/4) ) 18
- 19. (1/2) * ( 1 - (-1/4)n ) 12 earrings= --------------------------- ( 1 - (-1/4) ) Assume that there are R rings, P pins and E pair of ear-rings.= 1 / ( 2 * ( 1 - (-1/4) ) )= 2/5 It is given that, he has 2 1/2 times as many rings as pins.Similarly in East and West R = (5/2) * P or P = (2*R)/5directions= 1 - 1/4 + 1/16 - 1/64 + 1/256 - And, the number of pairs ofand so on... (a geometric series earrings is 4 less than thewith r = (-1/4) ) number of rings. E = R - 4 or R = E + 4 (1) * ( 1 - (-1/4)n )= --------------------------- Also, there are total 26 pieces. ( 1 - (-1/4) ) R + P + 2*E = 26 R + (2*R)/5 + 2*E = 26= 1 / ( ( 1- (-1/4) ) 5*R + 2*R + 10*E = 130= 4/5 7*R + 10*E = 130 7*(E + 4) + 10*E = 130So the soldier is 4/5 miles away 7*E + 28 + 10*E = 130towards East and 2/5 miles 17*E = 102away towards North. So using E=6right angled triangle, soldier is0.8944 miles away from his Hence, there are 6 pairs of Ear-camp. rings i.e. total 12 Ear-rings How many ways are there ofRaj has a jewel chest arranging the sixteen black orcontaining Rings, Pins and Ear- white pieces of a standardrings. The chest contains 26 international chess set on thepieces. Raj has 2 1/2 times as first two rows of the board?many rings as pins, and thenumber of pairs of earrings is 4 Given that each pawn isless than the number of rings. identical and each rook, knight and bishop is identical to itsHow many earrings does Raj pair.have? SubmittedAnswer Answer 19
- 20. 6,48,64,800 ways X He spent 1/3 for cloths =. (1/3)There are total 16 pieces which *Xcan be arranged on 16 places Remaining money = (2/3) * Xin 16P16 = 16! ways.(16! = 16 * 15 * 14 * 13 * 12 * He spent 1/5 of remaining..... * 3 * 2 * 1) money for food = (1/5) * (2/3) * X = (2/15) * XBut, there are some duplicate Remaining money = (2/3) * X -combinations because of (2/15) * X = (8/15) * Xidentical pieces. Again, he spent 1/4 of There are 8 identical remaining maoney for travel = pawn, which can be (1/4) * (8/15) * X = (2/15) * X arranged in 8P8 = 8! ways. Remaining money = (8/15) * X - Similarly there are 2 (2/15) * X = (6/15) * X identical rooks, 2 identical knights and 2 identical But after spending for travel he bishops. Each can be is left with Rs. 100/- So arranged in 2P2 = 2! ways. (6/15) * X = 100 X = 250Hence, the require answer is= (16!) / (8! * 2! * 2! * 2!) Grass in lawn grows equally= 6,48,64,800 thick and in a uniform rate. It takes 24 days for 70 cows and 60 days for 30 cows to eat theA person with some money whole of the grass.spends 1/3 for cloths, 1/5 of theremaining for food and 1/4 of How many cows are needed tothe remaining for travel. He is eat the grass in 96 days?left with Rs 100/- AnswerHow much did he have with him 20 cowsin the begining? g - grass at the beginningAnswer r - rate at which grass grows, per dayRs. 250/- y - rate at which one cow eats grass, per dayAssume that initially he had Rs. n - no of cows to eat the grass 20
- 21. in 96 days are (3, 0, 7) or (4, 1, 8) or (5, 2, 9)From given data,g + 24*r = 70 * 24 * y ---------- A It is given that there are 3 pairsg + 60*r = 30 * 60 * y ---------- B whose sum is 11. All possibleg + 96*r = n * 96 * y ---------- C pairs are (2, 9), (3, 8), (4, 7), (5, 6). Now required number is 5Solving for (B-A), digit number and it contains 3(60 * r) - (24 * r) = (30 * 60 * y) - pairs of 11. So it must not be(70 * 24 * y) having 0 and 1 in it. Hence, the36 * r = 120 * y ---------- D only possible combination for 2nd, 3rd and 4th digits is (5, 2,Solving for (C-B), 9)(96 * r) - (60 * r) = (n * 96 * y) -(30 * 60 * y) Also, 1st digit is thrice the last36 * r = (n * 96 - 30 * 60) * y digit. The possible120 * y = (n * 96 - 30 * 60) * y combinations are (3, 1), (6, 2)[From D] and (9, 3), out of which only (6,120 = (n * 96 - 1800) 2) with (5, 2, 9) gives 3 pairs ofn = 20 11. Hence, the answer is 65292.Hence, 20 cows are needed to Four friends - Arjan, Bhuvan,eat the grass in 96 days. Guran and Lakha wereThere is a safe with a 5 digit comparing the number of sheepnumber as the key. The 4th that they owned.digit is 4 greater than thesecond digit, while the 3rd digit It was found that Guran had tenis 3 less than the 2nd digit. The more sheep than Lakha.1st digit is thrice the last digit.There are 3 pairs whose sum is If Arjan gave one-third to11. Bhuvan, and Bhuvan gave a quarter of what he then held toFind the number. Guran, who then passed on aAnswer fifth of his holding to Lakha, they would all have an equal65292 number of sheep.As per given conditions, there How many sheep did each ofare three possible combinations them possess? Give thefor 2nd, 3rd and 4th digits. They minimal possible answer 21
- 22. Arjans sheep = Lakhas sheepAnswer 2A/3 = A/60 + B/20 + G/5 + L 2A/3 = A/60 + A/36 + 11A/90 + L (as B=5A/9 and G=11A/18) 2A/3 = A/6 + L A/2 = L A = 2L Also, it is given that Guran had ten more sheep than Lakha. G = L + 10 11A/18 = A/2 + 10 A/9 = 10 A = 90 sheepArjan, Bhuvan, Guran andLakha had 90, 50, 55 and 45 Thus, Arjan had 90 sheep,sheep respectively. Bhuvan had 5A/9 i.e. 50 sheep, Guran had 11A/18 i.e. 55Assume that Arjan, Bhuvan, sheep and Lakha had A/2 i.e.Guran and Lakha had A, B, G 45 sheep.and L sheep respectively. As it Consider a number 235, whereis given that at the end each last digit is the sum of first twowould have an equal number of digits i.e. 2 + 3 = 5.sheep, comparing the finalnumbers from the above table. How many such 3-digit numbers are there?Arjans sheep = Bhuvanssheep Answer2A/3 = A/4 + 3B/48A = 3A + 9B There are 45 different 3-digit5A = 9B numbers.Arjans sheep = Gurans sheep The last digit can not be 0.2A/3 = A/15 + B/5 + 4G/52A/3 = A/15 + A/9 + 4G/5 (as If the last digit is 1, the onlyB=5A/9) possible number is 101. (Note30A = 3A + 5A + 36G that 011 is not a 3-digit number)22A = 36G11A = 18G If the last digit is 2, the possible numbers are 202 and 112. 22
- 23. If the last digit is 3, the possible If its rightmost digit is placed atnumbers are 303, 213 and 123. its left end, then new number is 428571 which is 50% largerIf the last digit is 4, the possible than the original numbernumbers are 404, 314, 224 and 285714.134. The simplest way is to write aIf the last digit is 5, the possible small program. And the othernumbers are 505, 415, 325, way is trial and error !!!235 and 145. Two identical pack of cards A and B are shuffled throughly.Note the pattern here - If the One card is picked from A andlast digit is 1, there is only one shuffled with B. The top cardnumber. If the last digit is 2, from pack A is turned up. If thisthere are two numbers. If the is the Queen of Hearts, whatlast digit is 3, there are three are the chances that the topnumbers. If the last digit is 4, card in B will be the King ofthere are four numbers. If the Hearts?last digit is 5, there are fivenumbers. And so on..... AnswerThus, total numbers are 52 / 27031+2+3+4+5+6+7+8+9= 45 There are two cases to be considered.Altogether then, there are 45different 3-digit numbers, where CASE 1 : King of Hearts islast digit is the sum of first two drawn from Pack A anddigits. shuffled with Pack BFind the smallest number such Probability of drawing King ofthat if its rightmost digit is Hearts from Pack A = 1/51 (asplaced at its left end, the new Queen of Hearts is not to benumber so formed is precisely drawn)50% larger than the original Probability of having King ofnumber. Hearts on the top of the Pack BAnswer = 2/53The answer is 285714. So total probability of case 1 = 23
- 24. (1/51) * (2/53) = 2 / (51 * 53) 4. A->B, B->A, C->B 5. A->C, C->B, B->ACASE 2 : King of Hearts is not 6. A->C, C->B, B->Cdrawn from Pack A 7. A->C, C->A, B->A 8. A->C, C->A, B->CProbability of not drawing Kingof Hearts from Pack A = 50/51(as Queen of Hearts is not to Out of which, there are only twobe drawn) cases under which the antsProbability of having King of wont collide :Hearts on the top of the Pack B= 1/53 A->B, B->C, C->A A->C, C->B, B->ASo total probability of case 2 =(50/51) * (1/53) = 50 / (51 * 53) Find all sets of consecutiveNow adding both the integers that add up to 1000.probability, the required Submitted by : Jamesprobability is Barberousse= 2 / (51 * 53) + 50 / (51 * 53)= 52 / (51 * 53)= 52 / 2703 Answer= 0.0192378There are 3 ants at 3 corners of There are total 8 such series:a triangle, they randomly startmoving towards another corner. 1. Sum of 2000 numbers starting from -999 i.e.What is the probability that they summation of numbersdont collide? from -999 to 1000. (-999) + (-998) + (-997)Answer + ..... + (-1) + 0 + 1 + 2 + ..... + 997 + 998 +Lets mark the corners of the 999 + 1000 = 1000triangle as A,B,C. There are 2. Sum of 400 numberstotal 8 ways in which ants can starting from -197 i.e.move. summation of numbers from -197 to 202. 1. A->B, B->C, C->A (-197) + (-196) + (-195) 2. A->B, B->C, C->B + ..... + (-1) + 0 + 1 + 2 3. A->B, B->A, C->A 24
- 25. + ..... + 199 + 200 + summation of numbers 201 + 202 = 1000 from 198 to 202.3. Sum of 125 numbers 198 + 199 + 200 +201 starting from -54 i.e. + 202 = 1000 summation of numbers 8. Sum of 1 number from -54 to 70. starting from 1000. (-54) + (-53) + (-52) + 1000 = 1000 ..... + (-1) + 0 + 1 + 2 + ..... + 68 + 69 + 70 = 10004. Sum of 80 numbers starting from -27 i.e. There is a 4-character code, summation of numbers with 2 of them being letters and from -27 to 52. the other 2 being numbers. (-27) + (-26) + (-25) + ..... + (-1) + 0 + 1 + 2 + How many maximum attempts ..... + 50 + 51 + 52 = would be necessary to find the 1000 correct code? Note that the5. Sum of 25 numbers code is case-sensitive. starting from 28 i.e. Answer summation of numbers from 28 to 52. The maximum number of 28 + 29 + 30 + 31 + 32 attempts required are + 33 + 34 + 35 + 36 + 16,22,400 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + There are 52 possible letters - a 46 + 47 + 48 + 49 + 50 to z and A to Z, and 10 possible + 51 + 52 = 1000 numbers - 0 to 9. Now, 46. Sum of 16 numbers characters - 2 letters and 2 starting from 55 i.e. numbers, can be selected in summation of numbers 52*52*10*10 ways. These 4 from 55 to 70. characters can be arranged in 4 55 + 56 + 57 + 58 + 59 C2 i.e. 6 different ways - the +60 + 61 + 62 + 63 + number of unique patterns that 64 + 65 + 66 + 67 + 68 can be formed by lining up 4 + 69 + 70 = 1000 objects of which 2 are7. Sum of 5 numbers distinguished one way (i.e. they starting from 198 i.e. must be letters) and the other 2 are distinguished another way 25
- 26. (i.e. they must be numbers). first.Consider an example : Lets There are 8 corner cubes,assume that @ represents which can be arranged in 8!letter and # represents number. ways.the 6 possible ways of Each of these 8 cubes can bearranging them are : @@##, turned in 3 different directions,@#@#, @##@, #@@#, so there are 3^8 orientations#@#@, ##@@ altogether. But if you get all but one of the corner cube intoHence, the required answer is chosen positions and= 52*52*10*10*6 orientations, only one of 3= 16,22,400 attempts orientations of the final corner= 1.6 million approx. cube is possible. Thus, total ways corner cubes can beThanks to Tim Sanders for placed = (8!) * (3^8)/8 = (8!) *opening BrainVistas brain !!! (3^7)How many possiblecombinations are there in a Similarly, 12 edge cubes can3x3x3 rubics cube? be arranged in 12! ways. Each of these 12 cubes can beIn other words, if you wanted to turned in 2 different directions,solve the rubics cube by trying so there are 2^12 orientationsdifferent combinations, how altogether. But if you get all butmany might it take you (worst one of the edge cube intocase senerio)? chosen positions and orientations, only one of 2How many for a 4x4x4 cube? orientations of the final edgeSubmitted cube is possible. Thus, total ways edge cubes can beAnswer placed = (12!) * (2^12)/2 = (12!) * (2^11)There are 4.3252 * 10^19possible combinations for Here, we have essentially3x3x3 Rubics and 7.4012 * pulled the cubes apart and10^45 possible combinations stuck cubes back in placefor 4x4x4 Rubics. wherever we please. In reality, we can only move cubes around by turning the faces ofLets consider 3x3x3 Rubics the cubes. It turns out that you 26
- 27. cant turn the faces in such a N E Vway as to switch the positions E Rof two cubes while returning allthe others to their original L E Apositions. Thus if you get all but V Etwo cubes in place, there isonly one attainable choice for +them (not 2!). Hence, we must M Edivide by 2. -----------Total different possible ------combinations are= [(8!) * (3^7)] * [(12!) * (2^11)] / A L O2 N E= (8!) * (3^7) * (12!) * (2^10) Note that the leftmost letter= 4.3252 * 10^19 cant be zero in any word. Also, there must be a one-to-one mapping between digits andSimilarly, for 4x4x4 Rubics total letters. e.g. if you substitute 3different possible combinations for the letter M, no other letterare can be 3 and all other M in the= [(8!) * (3^7)] * [(24!)] * [(24!) / puzzle must be 3.(4!^6)] / 24= 7.4011968 * 10^45 AnswerNote that there are 24 edge A tough one!!!cubes, which you can not turnin 2 orientations (hence no Since R + E + E = 10 + E, it is2^24 / 2). Also, there are 4 clear that R + E = 10 andcenter cubes per face i.e. (24!) / neither R nor E is equal to 0 or(4!^6). You can switch 2 cubes 5. This is the only entry point towithout affecting the rest of thecombination as 4*4*4 has even solve it. Now use trial-n-errordimensions (hence no division method.by 2). But pattern on one side isrotated in 4 directions over 6faces, hence divide by 24.Substitute digits for the lettersto make the following relationtrue. 27
- 28. sex then the man is older N E V E R than the woman.2 1 4 1 9 5. If the dancer is a woman, then the dancer is older L E A V E than the singer.3 1 5 4 1 Whose occupation do you + M E know? And what is his/her+ 6 1 occupation? Answer ---------------------------------- Cindy is the Singer. Mr. Clinton or Monika is the A L O N E Dancer.5 3 0 2 1 From (1) and (3), the singer and the dancer, both can not beOne of the four people - Mr. a man. From (3) and (4), if theClinton, his wife Monika, their singer is a man, then theson Mandy and their daughter dancer must be a man. Hence,Cindy - is a singer and another the singer must be a woman.is a dancer. Mr. Clinton is olderthan his wife and Mady is older CASE I : Singer is a womanthan his sister. and Dancer is also a woman Then, the dancer is Monika and 1. If the singer and the the singer is Cindy. dancer are the same sex, then the dancer is older CASE II : Singer is a woman than the singer. and Dancer is also a man 2. If neither the singer nor Then, the dancer is Mr. Clinton the dancer is the parent and the singer is Cindy. of the other, then the singer is older than the In both the cases, we know that dancer. Cindy is the Singer. And either 3. If the singer is a man, Mr. Clinton or Monika is the then the singer and the Dancer. dancer are the same age. There are 20 people in your 4. If the singer and the applicant pool, including 5 pairs dancer are of opposite of identical twins. 28
- 29. If you hire 5 people randomly, 10C3 * what are the chances you will 3 2 10C2 3600 8/9 * 6/8 hire at least 1 pair of identical twins? (Needless to say, this 10C4 * could cause trouble ;)) 4 1 8/9 * 6/8 * 10C1 800 Submitted 4/7 10C5 * Answer 5 0 8/9 * 6/8 * 10C0 32 4/7 * 2/6 The probability to hire 5 Total 11584 people with at least 1 pair of identical twins is 25.28% Thus, total possible ways to 5 people from the 20 people hire 5 people without a single can be hired in 20C5 = 15504 pair of indentical twins = 11584 ways. ways Now, divide 20 people into two So, total possible ways to hire 5 groups of 10 people each : people with at least a single G1 - with all twins pair of indentical twins = 15504 G2 - with all people other than - 11584 = 3920 ways twins Hence, the probability to hire 5 Lets find out all possible ways people with at least a single to hire 5 people without a single pair of indentical twins pair of indentical twins. = 3920/15504 = 245/969People People No of No Total = 0.2528 from from ways to of ways = 25.28% G1 G2 hire G1 ways In a hotel, rooms are numbered without a to from 101 to 550. A room is single hire chosen at random. What is the pair of G2 probability that room number indentical starts with 1, 2 or 3 and ends twins with 4, 5 or 6? 0 5 10C0 10C5 252 1 4 10C1 10C4 2100 Answer 10C2 * 2 3 10C3 4800 There are total 450 rooms. 8/9 29
- 30. Out of which 299 room number Answerstarts with either 1, 2 or 3. (asroom number 100 is not there) One way to solve it is byNow out of those 299 rooms making 3 equations and solveonly 90 room numbers end with them simultaneously. But there4, 5 or 6 is rather easier way to solve it using Backtracing.So the probability is 90/450 i.e.1/5 or 0.20 Its given that at the end, eachDraw 9 dots on a page, in the had 24 tractors (24, 24, 24) i.e.shape of three rows of three after Z gave tractors to X & Ydots to form a square. Now as many as they had. It meansplace your pen on the page, that after getting tractors from Zdraw 4 straight lines and try their tractors got doubled. Soand cover all the dots. before Z gave them tractors, they had 12 tractors each and ZYoure not allowed to lift your had 48 tractors. (12, 12, 48)pen. Similarly, before Y gaveNote: Dont be confined by the tractors to X & Z, they had 6 &dimensions of the square. 24 tractors respectively and YSubmitted had 42 tractors i.e. (6, 42, 24) Again, before X gave tractors to Y & Z, they had 21 & 12 tractors respectively and X had 39 tractors i.e. (39, 21, 12)There are 3 persons X, Y andZ. On some day, X lent tractors Hence, initially X had 39to Y and Z as many as they tractors, Y had 21 tractorshad. After a month Y gave as and Z had 12 tractors.many tractors to X and Z asmany as they have. After a There is a 50m long armymonth Z did the same thing. At platoon marching ahead. Thethe end of this transaction each last person in the platoon wantsone of them had 24. to give a letter to the first person leading the platoon. SoFind the tractors each originally while the platoon is marchinghad? he runs ahead, reaches the first person and hands over the 30
- 31. letter to him and without the last person moved [50-(50-stopping he runs and comes X)] X meters whereas theback to his original position. platoon moved (50-X) meters.In the mean time the whole Now, as the ratios are equal,platoon has moved ahead by (50+X)/X = X/(50-X)50m. (50+X)*(50-X) = X*XThe question is how much Solving, X=35.355 metersdistance did the last personcover in that time. Assuming Thus, total distance covered bythat he ran the whole distance the last personwith uniform speed. = (50+X) + XSubmitted = 2*X + 50 = 2*(35.355) + 50Answer = 120.71 metersThe last person covered120.71 meters. Note that at first glance, one might think that the totalIt is given that the platoon and distance covered by the lastthe last person moved with person is 100 meters, as he ranuniform speed. Also, they both the total lenght of the platoonmoved for the identical amount (50 meters) twice. TRUE, butof time. Hence, the ratio of the thats the relative distancedistance they covered - while covered by the last person i.e.person moving forward and assuming that the platoon isbackword - are equal. stationary. Assume that you have enoughLets assume that when the last coins of 1, 5, 10, 25 and 50person reached the first person, cents.the platoon moved X metersforward. How many ways are there to make change for a dollar? DoThus, while moving forward the explain your answer.last person moved (50+X)meters whereas the platoonmoved X meters.Similarly, while moving back 31
- 32. waysThere are 292 ways to makechange for a dollar using Lets take another example:coins of 1, 5, 10, 25 and 50 To get change for 75 centscents. using all coins up to 50 cent i.e. 1, 5, 10, 25 and 50 cents coins.Lets generalised the teaser * 75 cents change using coinsand make a table as shown upto 25 cent = 121 waysabove. * (75-50) 25 cents change using coins upto 50 cent = 13If you wish to make change for ways75 cents using only 1, 5, 10 and * 75 cents change using coins25 cent coins, go to the .25 row upto 50 cent = 121+13 = 134and the 75 column to obtain ways121 ways to do this. For people who dont want toThe table can be created from tease their brain and love to doleft-to-right and top-to-bottom. computer programming, thereStart with the top left i.e. 1 cent is a simple way. Write a smallrow. There is exactly one way multi-loop program to solve theto make change for every equation: A + 5B + 10C + 25Damount. Then calculate the 5 + 50E = 100cents row by adding the where,number of ways to make A = 0 to 100change for the amount using 1 B = 0 to 20cent coins plus the number of C = 0 to 10ways to make change for 5 D = 0 to 4cents less using 1 and 5 cent E = 0 to 2coins. The program should output allLets take an example: the possible values of A, B, C,To get change for 50 cents D and E for which the equationusing 1, 5 and 10 cent coins. is satisfied.* 50 cents change using 1 and5 cent coins = 11 ways In a Road Race, one of the* (50-10) 40 cents change three bikers was doing 15kmusing 1, 5 and 10 cent coins = less than the first and 3km25 ways more than the third. He also* 50 cents change using 1, 5 finished the race 12 minutesand 10 cent coins = 11+25 = 36 after the first and 3 minutes 32
- 33. before the third. Now as per the data given in the teaser, at a time T minCan you find out the speed of X1 = V1 * Teach biker, the time taken by ----> 1each biker to finish the raceand the length of the course? X1 - 15 = V2 * T ----> 2Assume that there were nostops in the race and also they X1 - 18 = V3 * Twere driving with constant ----> 3speeds through out the At a Distance S Km.Answer S = V1 * T1 ----> 4 S = V2 * (T1 + 12) ----> 5 S = V3 * (T1 + 15) ----> 6 Thus there are 6 equations and 7 unknown data that means it has infinite number of solutions. By solving above 6 equations we get, Time taken by first biker, T1 =Let us assume that 60 Min.Speed of First biker = V1 Time taken by Second biker, T2km/min = 72 Min.Speed of Second biker = V2 Time taken by first biker, T3 =km/min 75 Min.Speed of Third biker = V3km/min Also, we getTotal time take by first biker = Speed of first biker, V1 = 90/TT1 min km/minTotal distance = S km Speed of second biker, V2 = (5/6)V1 = 75/T km/min Speed of third biker, V3 = 33
- 34. (4/5)V1 = 72/T km/min It is clear that option 3 is not possible. So we are left withAlso, the length of the course, only two options. Also, the lastS = 5400/T km digit is three times the second, which rules out the secondThus, for the data given, only option. Hence, the answer isthe time taken by each biker 1349.can be found i.e. 60, 72 and 75minutes. For other quantities, Difference between Bholus andone more independent datum is Molus age is 2 years and therequired i.e. either T or V1 or difference between Molus andV2 or V3 Kolus age is 5 years.Thanks to Theertham Srinivas What is the maximum possiblefor the answer !!! value of the sum of theWhat is the four-digit number in difference in their ages, takenwhich the first digit is 1/3 of the two at a time?second, the third is the sum of Answerthe first and second, and thelast is three times the second? The maximum possible value of the sum of the differenceAnswer in their ages - taken two at a time - is 14 years.The 4 digit number is 1349. It is given that -It is given that the first digit is "Difference between Bholus1/3 of the second. There are 3 and Molus age is 2 years"such possibilities. "Difference between Molus and Kolus age is 5 years" 1. 1 and 3 2. 2 and 6 Now, to get the maximum 3. 3 and 9 possible value, the difference between Bholus and KolusNow, the third digit is the sum age should be maximum i.e.of the first and second digits. Molus age should be in between Bholus and Kolus 1. 1 + 3 = 4 age. Then, the difference 2. 2 + 6 = 8 between Bholus and Kolus 3. 3 + 9 = 12 age is 7 years. 34
- 35. Hence, the maximum possible (2) 5 - 2 = 3value of the sum of the 10 (0) x 2 = 20difference in their ages - taken (3) 6 / 3 = 2two at a time - is (2 + 5 + 7) 14 14 (4) - 3 = 11years.If it is given that: Answer 3 : 1425 - 2 = 3 Drop left and right digit100 x 2 = 20 alternatively from the actual36 / 3 = 2 answer. 25 - 2 = (2) 3 (drop left digit i.e.What is 144 - 3 = ? 2)Submitted 100 * 2 = 20 (0) (drop right digitAnswer i.e. 0) 36 / 3 = (1) 2 (drop left digit i.e.There are 3 possible answers 1)to it. 144 - 3 = 14 (1) (drop right digit i.e. 1)Answer 1 : 9Simply replace the first numberby its square root. A 3 digit number is such that its(25) 5 - 2 = 3 unit digit is equal to the product(100) 10 x 2 = 20 of the other two digits which are(36) 6 / 3 = 2 prime. Also, the difference(144) 12 - 3 = 9 between its reverse and itself is 396.Answer 2 : 11Drop the digit in the tens What is the sum of the threeposition from the first number. digits?(2) 5 - 2 = 31 (0) 0 x 2 = 20 Answer(3) 6 / 3 = 21 (4) 4 - 3 = 11 The required number is 236 and the sum is 11.You will get the same answeron removing left and right digit It is given that the first twoalternatively from the first digits of the required numbernumber i.e remove left digit are prime numbers i.e. 2, 3, 5from first (2), right digit from or 7. Note that 1 is neithersecond (0), left digit from third prime nor composite. Also, the(3) and right digit from forth (4). third digit is the multiplication of 35
- 36. the first two digits. Thus, first Hence, there are 4 marblestwo digits must be either 2 or 3 under each mug.i.e. 22, 23, 32 or 33 which At University of Probability,means that there are four there are 375 freshmen, 293possible numbers - 224, 236, sophomores, 187 juniors, & 126326 and 339. seniors. One student will randomly be chosen to receiveNow, it is also given that - the an award.difference between its reverseand itself is 396. Only 236 What percent chance is theresatisfies this condition. Hence, that it will be a junior? Round tothe sum of the three digits is the nearest whole percent11. AnswerThere are 4 mugs placed 19%upturned on the table. Eachmug have the same number of This puzzle is easy. Divide themarbles and a statement about number of juniors (187) by thethe number of marbles in it. The total number of students (981),statements are: Two or Three, & then multiply the number byOne or Four, Three or One, 100 to convert to a percentage.One or Two. Hence the answer isOnly one of the statement is (187/981)*100 = 19%correct. How many marbles arethere under each mug? If you were to dial any 7 digitsAnswer on a telephone in random order, what is the probabilityA simple one. that you will dial your own phone number?As it is given that only one ofthe four statement is correct, Assume that your telephonethe correct number can not number is 7-digits.appear in more than one Answerstatement. If it appears in morethan one statement, then more 1 in 10,000,000than one statement will becorrect. There are 10 digits i.e. 0-9. First digit can be dialed in 10 36
- 37. ways. Second digit can be words.dialed in 10 ways. Third digitcan be dialed in 10 ways. And It is a simple permutationso on..... problem of arranging 6 letters to get different six-letter words.Thus, 7-digit can be dialed in And it can be done in in 6! ways10*10*10*10*10*10*10 i.e. 720 ways.(=10,000,000) ways. And, youhave just one telephone In otherwords, the first letternumber. Hence, the possibility can be any of the given 6that you will dial your own letters (A through F). Then,number is 1 in 10,000,000. whatever the first letter is, the second letter will always beNote that 0123456 may not be from the remaining 5 letters (asa valid 7-digit telephone same letter can not be usednumber. But while dialing in twice), and the third letterrandom order, that is one of the always be from the remaining 4possible 7-digit number which letters, and so on. Thus, theyou may dial. different possible six-letterAn anthropologist discovers an words are 6*5*4*3*2*1 = 720isolated tribe whose writtenalphabet contains only six Kate, Demi, Madona, Sharon,letters (call the letters A, B, C, Britney and Nicole decided toD, E and F). The tribe has a lunch together in a restaurant.taboo against using the same The waiter led them to a roundletter twice in the same word. table with six chairs.Its never done. How many different ways canIf each different sequence of they seat?letters constitues a differentword in the language, what is Answerthe maximum number of six-letter words that the language There are 120 differentcan employ? possible seatingSubmitted arrangments.Answer Note that on a round table ABCDEF and BCDEFA is theThe language can employ same.maximum of 720 six-letter 37
- 38. The first person can sit on any = 1 / 372one of the seats. Now, for the = 0.002688second person there are 5options, for the third person What is the area of the trianglethere are 4 options, for the forth ABC with A(e,p) B(2e,3p) andperson there are 3 options, for C(3e,5p)?the fifth person there are 2options and for the last person where p = PI (3.141592654)there is just one option. AnswerThus, total different possible A tricky ONE.seating arrangements are=5*4*3*2*1 Given 3 points are colinear.= 120 Hence, it is a straight line.3 blocks are chosen randomly on achessboard. What is the probability Hence area of triangle is 0.that they are in the same diagonalAnswer Silu and Meenu were walkingThere are total of 64 blocks on on the road.a chessboard. So 3 blocks canbe chosen out of 64 in 64C3 Silu said, "I weigh 51 Kgs. Howways. much do you weigh?"So the sample space is =41664 Meenu replied that she wouldnt reveal her weight directly asThere are 2 diagonal on she is overweight. But she said,chessboard each one having 8 "I weigh 29 Kgs plus half of myblocks. Consider one of them. weight."3 blocks out of 8 blocks indiagonal can be chosen in 8C3 How much does Meenu weigh?ways.But there are 2 such diagonals, Answerhence favourables = 2 * 8C3 = 2* 56 = 112 Meenu weighs 58 Kgs.The require probability is It is given that Meenu weighs= 112 / 41664 29 Kgs plus half of her own weight. It means that 29 Kgs is 38
- 39. the other half. So she weighs leaves a remainder of 0, 3, or 658 Kgs. if you remove it from the sum 0 + 1 + ... + 9. Hence, it followsSolving mathematically, lets that J must be 9.assume that her weight is X A man has Ten Horses andKgs. nine stables as shown here.X = 29 + X/2 [] [] [] [] [] [] []2*X = 58 + X [] []X = 58 Kgs The man wants to fit Ten Horses into nine stables. HowConsider the sum: ABC + DEF can he fit Ten horses into nine+ GHI = JJJ stables? SubmittedIf different letters represent Answerdifferent digits, and there are noleading zeros, what does J The answer is simple. It saysrepresent? the man wants to fit "Ten Horses" into nine stables. There are nine letters in theAnswer phrase "Ten Horses". So you can put one letter each in allThe value of J must be 9. nine stables. [T] [E] [N] [H] [O]Since there are no leading [R] [S] [E] [S]zeros, J must be 7, 8, or 9. (JJJ= ABC + DEF + GHI = 14? + A man is at a river with a 925? + 36? = 7??) gallon bucket and a 4 gallon bucket. He needs exactly 6Now, the remainder left after gallons of water.dividing any number by 9 is thesame as the remainder left after How can he use both bucketsdividing the sum of the digits of to get exactly 6 gallons ofthat number by 9. Also, note water?that 0 + 1 + ... + 9 has aremainder of 0 after dividing by Note that he cannot estimate by9 and JJJ has a remainder of 0, dumping some of the water out3, or 6. of the 9 gallon bucket or the 4 gallon bucketThe number 9 is the only Answernumber from 7, 8 and 9 that 39
- 40. For the sack of explanation, with 9lets identify 4 gallon bucket as gallonBucket P and 9 gallon bucket wateras Bucket Q. Pour 3 4 gallon 9 gallon gallon bucket bucket water from 4 6Operation (Bucket (Bucket bucket Q to P) Q) bucket PInitially 0 0Fill the 9 gallon bucket contains 6bucket Q gallon of water, as required.with 9 0 9gallon Each of the five characters inwater the word BRAIN has a different value between 0 and 9. UsingPour 4 the given grid, can you find outgallon the value of each character?water from 4 5 B R A Ibucket Q to N 31bucket P B B R BEmpty 0 5 A 31bucket PPour 4 N I A Bgallon B 32water from 4 1bucket Q to N I B Abucket P I 30Empty I R A A 0 1bucket P A 23Pour 1gallon 37 29 25 27water from 1 0 29bucket Q to The numbers on the extremebucket P right represent the sum of the values represented by theFill the characters in that row. Also, the 1 9bucket Q numbers on the last raw 40
- 41. represent the sum of the values B + R + A + I + N = 31represented by the characters 7 + 6 + 4 + 5 + N = 31in that column. e.g. B + R + A + N=9I + N = 31 (from first row)Answer Thus, B=7, R=6, A=4, I=5 and N=9B=7, R=6, A=4, I=5 and N=9 There are 9 coins. Out of which one is odd one i.e weight is lessMake total 10 equations - 5 for or more. How many iterationsrows and 5 for columns - and of weighing are required to findsovle them. odd coin? AnswerFrom Row3 and Row4,N+I+A+B+B=N+I+B+ It is always possible to find oddA+I+2 coin in 3 weighings and to tellB=I+2 whether the odd coin is heavier or lighter.From Row1 and Row3,B+R+A+I+N=N+I+A+ 1. Take 8 coins and weigh 4B+B-1 against 4.R=B-1 o If both are not equal, goto step 2From Column2, o If both are equal,R + B + I + I + R = 29 goto step 3B + 2R + 2I = 29B + 2(B - 1) + 2I = 293B + 2I = 313(I + 2) + 2I = 31 2. One of these 8 coins is5I = 25 the odd one. Name theI=5 coins on heavier side of the scale as H1, H2, H3Hence, B=7 and R=6 and H4. Similarly, name the coins on the lighterFrom Row2, side of the scale as L1,B + B + R + B + A = 31 L2, L3 and L4. Either one3B + R + A = 31 of Hs is heavier or one of3(7) + 6 + A = 31 Ls is lighter. Weigh (H1,A=4 H2, L1) against (H3, H4, X) where X is one coinFrom Row1, 41
- 42. remaining in intialweighing. o If (H3, H4, X) is o If both are equal, heavier side on the one of L2, L3, L4 is scale, either H3 or lighter. Weigh L2 H4 is heavier or L1 against L3. is lighter. Weight H3 If both are against H4 equal, L4 is If both are the odd coin equal, L1 is and is lighter. the odd coin If L2 is light, and is lighter. L2 is the odd If H3 is heavy, coin and is H3 is the odd lighter. coin and is If L3 is light, heavier. L3 is the odd If H4 is heavy, coin and is H4 is the odd lighter. coin and is heavier. o If (H1, H2, L1) is heavier side on the 3. The remaining coin X is scale, either H1 or the odd one. Weigh X H2 is heavier. against the anyone coin Weight H1 against used in initial weighing. H2 o If both are equal, If both are there is some error. equal, there is o If X is heavy, X is some error. the odd coin and is If H1 is heavy, heavier. H1 is the odd o If X is light, X is the coin and is odd coin and is heavier. lighter. If H2 is heavy, H2 is the odd In a sports contest there were coin and is m medals awarded on n heavier. successive days (n > 1). 42
- 43. 1. On the first day 1 medal any other simpler method, do and 1/7 of the remaining submit it. m - 1 medals were awarded. A number of 9 digits has the 2. On the second day 2 following properties: medals and 1/7 of the now remaining medals The number comprising was awarded; and so on. the leftmost two digits is 3. On the nth and last day, divisible by 2, that the remaining n medals comprising the leftmost were awarded. three digits is divisible by 3, the leftmost four by 4,How many days did the contest the leftmost five by 5, andlast, and how many medals so on for the nine digits ofwere awarded altogether? the number i.e. theAnswer number formed from the first n digits is divisible byTotal 36 medals were n, 2<=n<=9.awarded and the contest was Each digit in the numberfor 6 days. is different i.e. no digits are repeated.On day 1: Medals awarded = (1 The digit 0 does not occur+ 35/7) = 6 : Remaining 30 in the number i.e. it ismedals comprised only of theOn day 2: Medals awarded = (2 digits 1-9 in some order.+ 28/7) = 6 : Remaining 24medals Find the number.On day 3: Medals awarded = (3+ 21/7) = 6 : Remaining 18 AnswermedalsOn day 4: Medals awarded = (4 The answer is 381654729+ 14/7) = 6 : Remaining 12medals One way to solve it is Trial-&-On day 5: Medals awarded = (5 Error. You can make it bit+7/7) = 6 : Remaining 6 medals easier as odd positions willOn day 6: Medals awarded 6 always occupy ODD numbers and even positions will alwaysI got this answer by writing occupy EVEN numbers. Furthersmall program. If anyone know 5th position will contain 5 as 0 does not occur. 43
- 44. You have to figure out who isThe other way to solve this who IN ONLY 2 QUESTIONS.problem is by writing a Your questions have to be YEScomputer program that or NO questions and can onlysystematically tries all be answered by one person. (Ifpossibilities. you ask the same question to two different people then that counts as two questions). Keep1/3 rd of the contents of a in mind that all four know eachcontainer evaporated on the 1st others characteristics whetherday. 3/4th of the remaining they lie or not.contents of the containerevaporated on the second day. What questions would you ask to figure out who is who?What part of the contents of the Remember that you can askcontainer is left at the end of only 2 questions.the second day? SubmittedAnswer You have 3 baskets, & each one contains exactly 4 balls,Assume that contents of the each of which is of the samecontainer is X size. Each ball is either red, black, white, or purple, & thereOn the first day 1/3rd is is one of each color in eachevaporated. basket.(1 - 1/3) of X is remaining i.e.(2/3)X If you were blindfolded, & lightly shook each basket so that theOn the Second day 3/4th is balls would be randomlyevaporated. Hence, distributed, & then took 1 ball(1- 3/4) of (2/3)X is remaining from each basket, what chancei.e. (1/4)(2/3)X = (1/6) X is there that you would have exactly 2 red balls?Hence 1/6th of the contents ofthe container is remaining AnswerThere are four people in a room There are 64 different possible(not including you). Exactly two outcomes, & in 9 of these,of these four always tell the exactly 2 of the balls will be red.truth. The other two always lie. There is thus a slightly better than 14% chance [(9/64)*100] 44
- 45. that exactly 2 balls will be red. Commission chooses 11 from those 80 numbers, again noA much faster way to solve the repetition. You win the lottery ifproblem is to look at it this way. atleast 7 of your numbers areThere are 3 scenarios where there in the 11 chosen by theexactly 3 balls are red: Lottery Commission. 1 2 3 What is the probablity of winning the lottery? ----------- Answer R R X The probability of winning the lottery is two in one billion i.e. R X R only two person can win from one billion !!! X R R Lets find out sample spaceX is any ball that is first. The Lottery Commissionnot red. chooses 11 numbers from theThere is a 4.6875% chance that 80. Hence, the 11 numberseach of these situations will from the 80 can be selected inoccur. 80 C11 ways which is very very high and is equal to 1.04776 *Take the first one, for example: 101325% chance the first ball is red,multiplied by a 25% chance the Now, you have to select 8second ball is red, multiplied by numbers from 80 which can bea 75% chance the third ball is selected in 80C8 ways. But wenot red. are interested in only those numbers which are in 11Because there are 3 scenarios numbers selected by thewhere this outcome occurs, you Lottery Commision. There are 2multiply the 4.6875% chance of cases.any one occurring by 3, & youget 14.0625% You might select 8 numbers which all areConsider a state lottery where there in 11 numbersyou get to choose 8 numbers choosen by the Lotteryfrom 1 to 80, no repetiton Commission. So there are 11allowed. The Lottery C8 ways. 45

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